In $\mathbb{R}^2$ are considered the lines $r: y=0$, $s: y=1$, $t: x=0$, their union $M=r\cup s\cup t$ and the line segment $A: x=0, 0\leq y \leq 1$. In $M$ is defined the equivalence relation $\mathfrak{R}$ that identifies all points of $A$. Now, $\mathbb{R}^2$ is equipped with the usual topology $\tau_u$ and $M/\mathfrak{R}$ with the quotient topology. Let $$p: M \to M/\mathfrak{R}$$ the canonical map. Show whether $p(t)$ is a neighbourhood of $p(A)$ in $M/\mathfrak{R}$ or not, and find a subset $E\subset M$ such $p(E)$ is.
This is the first exercise I try on quotient topology, and I'm struggling with the concepts required to solve it. First, I'm not sure what "identifying" all points of $A$ means. I know that if you identify two points, or three points with another one, then I could just consider all those three points to be the same I identified them with. But I don't know if identifying all points of $A$ means that there is a point $a$ which is equivalent to all the points $b\in A$.
Secondly, the set $M$ shares with $A$ all $A$. Does that mean that the canonical map $p$ on the points that are not in $A$ is the identity map?
If someone could clarify to me that questions and help me, I'd be very grateful. Thanks in advance!
Formally, the equivalence relation is given by $$\forall a,b\in\Bbb R^2,\; a\mathfrak R b \iff a\in A\text{ and }b\in A.$$ Thus all points of $A$ are identified (i.e. we declare that all points of $A$ are equal in the quotient) : by transitivity of $\mathfrak R$, this is equivalent to identifying all the points of $A$ to a single $a\in A$.
Finally, if $b\in M\setminus A$, then the equivalence class of $b$ contains only $b$ itself. But strictly speaking, the projection $p$ is not the identity on $M\setminus A$, simply because $M/\mathfrak R$ is not the same set as $M$. But intuitively, $p$ doesn't change $M\setminus A$ and contracts $A$ onto one single point.